Ô­×ÓÐòÊýÒÀ´ÎÔö´óµÄ¶ÌÖÜÆÚÔªËØa¡¢b¡¢c¡¢dºÍeÖУ¬aµÄ×îÍâ²ãµç×ÓÊýΪÆäÖÜÆÚÊýµÄ¶þ±¶£»bºÍdµÄA2BÐÍÇ⻯Îï¾ùΪVÐηÖ×Ó£¬cµÄ£«1¼ÛÀë×Ó±ÈeµÄ£­1¼ÛÀë×ÓÉÙ8¸öµç×Ó¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÔªËØaΪ________£¬cΪ________¡£

(2)ÓÉÕâЩԪËØÐγɵÄË«Ô­×Ó·Ö×ÓΪ________¡£

(3)ÓÉÕâЩԪËØÐγɵÄÈýÔ­×Ó·Ö×ÓÖУ¬·Ö×ӵĿռä½á¹¹ÊôÓÚÖ±ÏßÐεÄÊÇ________£¬·ÇÖ±ÏßÐεÄÊÇ________(д2ÖÖ)¡£

(4)ÕâЩԪËصĵ¥ÖÊ»òÓÉËüÃÇÐγɵÄABÐÍ»¯ºÏÎïÖУ¬Æ侧ÌåÀàÐÍÊôÓÚÔ­×Ó¾§ÌåµÄÊÇ________£¬Àë×Ó¾§ÌåµÄÊÇ________£¬½ðÊô¾§ÌåµÄÊÇ________£¬·Ö×Ó¾§ÌåµÄÊÇ________(ÿ¿ÕÌîÒ»ÖÖ)¡£

(5)ÔªËØaºÍbÐγɵÄÒ»ÖÖ»¯ºÏÎïÓëcºÍbÐγɵÄÒ»ÖÖ»¯ºÏÎï·¢ÉúµÄ·´Ó¦³£ÓÃÓÚ·À¶¾Ãæ¾ßÖУ¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____________________¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


»¯Ñ§ÓÃÓïÊÇѧϰ»¯Ñ§µÄÖØÒª¹¤¾ß£¬ÏÂÁл¯Ñ§ÓÃÓïÖУ¬ÕýÈ·µÄÊÇ

A£®NH4£«µÄË®½â·´Ó¦Àë×Ó·½³Ìʽ£ºNH4£« + H2O NH3·H2O + H£«

B£®ÔÚAgClÐü×ÇÒºÖмÓÈëKIÈÜÒº³ä·ÖÕñµ´£ºAg+ + I- = AgI¡ý

C£®¶Æͭʱ£¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª£º Cu2++2e£­ == Cu

D£®Ä³·´Ó¦⊿H£¼0,Ôò¸Ã·´Ó¦ÔÚÈÎÒâÌõ¼þϾù¿É×Ô·¢½øÐС£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


    ÓÉÓÚ¹¤Òµ·¢Õ¹£¬È¼ÉÕÉ豸²»¶ÏÔö¶à£¬É豸¹æÄ£²»¶ÏÔö´ó¡£ÕâЩ¹¤³§ÅŷŵÄÑÌÆøÖж¼»áÓдóÁ¿µÄSO2·ÏÆø¡£¾Ýͳ¼Æ£¬ÎÒ¹ú1995Ä깤ҵSO2µÄÅÅ·ÅÁ¿Îª1 396Íò¶Ö£¬2006Ä깤ҵSO2µÄÅÅ·ÅÁ¿´ïµ½ÁË3 800Íò¶Ö£¬ÓÉÓÚSO2µÄÎÛȾ£¬ÎÒ¹úÿÄêËðʧ¸ß´ï1 100ÒÚÔª¡£

(1)д³öÁ½¸ö¹¤ÒµÉú²ú¹ý³ÌÖвúÉúSO2µÄʵÀý£º

¢Ù________________________________________________________________________¡¢

¢Ú________________________________________________________________________¡£

(2)·ÀÖÎSO2ÎÛȾ¿É²ÉÓõĴëÊ©ÓÐ(д³öÈýÖÖ)£º

¢Ù________________________________________________________________________¡¢

¢Ú________________________________________________________________________¡¢

¢Û________________________________________________________________________¡£

(3)ʪʽʯ»Òʯ—ʯ¸à·¨ÍÑÁò¹¤ÒÕÊÇÑÌÆøÍÑÁò¼¼ÊõÖÐ×î³ÉÊìµÄÒ»ÖÖ·½·¨¡£Æ乤ÒÕÁ÷³ÌÊÇ£ºÑÌÆø¾­¹ø¯ԤÈÈÆ÷³öÀ´£¬½øÈëµç³ý³¾Æ÷³ýµô´ó²¿·Ö·Ûú»ÒÑ̳¾£¬ÔÙ¾­¹ýÒ»¸öרÃŵÄÈȽ»»»Æ÷£¬È»ºó½øÈëÎüÊÕËþ£¬ÑÌÆøÖеÄSO2Ó뺬ÓÐʯ»ÒʯµÄ½¬Òº½øÐÐÆøÒº½Ó´¥£¬Í¨Èë¿ÕÆøºóÉú³Éʯ¸à(CaSO4·2H2O)£¬¾­ÍÑÁòµÄÑÌÆø£¬Ó¦ÓÃÑ­»·ÆøÌå¼ÓÈÈÆ÷½øÐÐÔÙ¼ÓÈÈ£¬½øÈëÑÌ´Ñ£¬ÅÅÈë´óÆø¡£

¢Ùд³öʪ·¨Ê¯»Òʯ—ʯ¸à·¨ÍÑÁòËùÉæ¼°µÄ»¯Ñ§·´Ó¦·½³Ìʽ£º

________________________________________________________________________

________________________________________________________________________¡£

¢ÚÓÃʯ»Òʯ½¬Òº×÷SO2ÎüÊÕ¼Á¶ø²»ÓÃÊìʯ»ÒÎüÊÕSO2µÄÔ­ÒòÊÇ£º________________________________________________________________________

________________________________________________________________________¡£

¢ÛÉÏÊö¹ý³ÌÖеõ½µÄʯ¸à£¬Èç¹ûº¬ÂÈ»¯ºÏÎï(Ö÷ÒªÀ´Ô´ÓÚȼÁÏú)³¬¹ýÔÓÖʼ«ÏÞÖµ£¬Ôòʯ¸à²úÆ·ÐÔÄܱ仵¡£¹¤ÒµÉÏÏû³ý¿ÉÈÜÐÔÂÈ»¯ÎïµÄ·½·¨ÊÇ__________________¡£

(4)ij»¯Ñ§ÐËȤС×éΪÁ˲ⶨÑÌÆøÍÑÁòËùµÃʯ¸àµÄ×é³É(CaSO4·xH2O)¼´²â¶¨xÖµ£¬×öÈçÏÂʵÑ飺½«Ê¯¸à¼ÓÈÈʹ֮ÍÑË®£¬¼ÓÈȹý³ÌÖйÌÌåµÄÖÊÁ¿Óëʱ¼äµÄ±ä»¯¹ØϵÈçÏÂͼËùʾ¡£Êý¾Ý±íÃ÷µ±¹ÌÌåµÄÖÊÁ¿Îª2.72 gºó²»Ôٸı䡣Çó£º

¢Ùʯ¸àµÄ»¯Ñ§Ê½¡£¢ÚͼÏñÖÐAB¶Î¶ÔÓ¦»¯ºÏÎïµÄ»¯Ñ§Ê½¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¹èÔڵؿÇÖеĺ¬Á¿½Ï¸ß¡£¹è¼°Æ仯ºÏÎïµÄ¿ª·¢ÓÉÀ´ÒѾã¬ÔÚÏÖ´úÉú»îÖÐÓй㷺ӦÓ᣻شðÏÂÁÐÎÊÌ⣺

£¨1£©1810ÄêÈðµä»¯Ñ§¼Ò±´²ÉÀûÎÚ˹ÔÚ¼ÓÈÈʯӢɰ¡¢Ä¾Ì¿ºÍÌúʱ£¬µÃµ½Ò»ÖÖ¡°½ðÊô¡±¡£ÕâÖÖ¡°½ðÊô¡±¿ÉÄÜÊÇ        ¡£

£¨2£©ÌÕ´É¡¢Ë®ÄàºÍ²£Á§Êdz£ÓõĹèËáÑβÄÁÏ¡£ÆäÖУ¬Éú²úÆÕͨ²£Á§µÄÖ÷ÒªÔ­ÁÏÓР       ¡£

£¨3£©¸ß´¿¹èÊÇÏÖ´úÐÅÏ¢¡¢°ëµ¼ÌåºÍ¹â·ü·¢µçµÈ²úÒµ¶¼ÐèÒªµÄ»ù´¡²ÄÁÏ¡£¹¤ÒµÉÏÌá´¿¹èÓжàÖÖ·Ïߣ¬ÆäÖÐÒ»ÖÖ¹¤ÒÕÁ÷³ÌʾÒâͼ¼°Ö÷Òª·´Ó¦ÈçÏ£º

¢ÙÓÃʯӢɰºÍ½¹Ì¿Ôڵ绡¯ÖиßμÓÈÈÒ²¿ÉÒÔÉú²ú̼»¯¹è£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ        £»Ì¼»¯¹èÓֳƠ       £¬Æ侧Ìå½á¹¹Óë        ÏàËÆ¡£

¢ÚÔÚÁ÷»¯´²·´Ó¦µÄ²úÎïÖУ¬SiHCl3´óÔ¼Õ¼85%£¬»¹ÓÐSiCl4¡¢SiH2Cl2¡¢SiH3ClµÈ£¬ÓйØÎïÖʵķеãÊý¾ÝÈçÏÂ±í£¬Ìá´¿SiHCl3µÄÖ÷Òª¹¤ÒÕ²Ù×÷ÒÀ´ÎÊdzÁ½µ¡¢ÀäÄýºÍ        ¡£

ÎïÖÊ

Si

SiCl4

SiHCl3

SiH2Cl2

SiH3Cl

HCl

SiH4

·Ðµã/¡æ

2355

57.6

31.8

8.2

-30.4

-84.9

-111.9

¢ÛSiHCl3¼«Ò×Ë®½â£¬ÆäÍêÈ«Ë®½âµÄ²úÎïΪ        ¡£

£¨4£©Âȼҵ¿ÉΪÉÏÊö¹¤ÒÕÉú²úÌṩ²¿·ÖÔ­ÁÏ£¬ÕâЩԭÁÏÊÇ        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


    ij¹¤³§ÓÃÈíÃÌ¿ó(º¬MnO2Ô¼70%¼°Al2O3)ºÍÉÁп¿ó(º¬ZnSÔ¼80%¼°FeS)¹²Í¬Éú²úMnO2ºÍZn(¸Éµç³ØÔ­ÁÏ)¡£

ÒÑÖª£º¢ÙAÊÇMnSO4¡¢ZnSO4¡¢Fe2(SO4)3¡¢Al2(SO4)3µÄ»ìºÏÒº¡£

¢Ú¢ôÖеç½â·½³ÌʽΪMnSO4£«ZnSO4£«2H2OMnO2£«Zn£«2H2SO4¡£

(1)AÖÐÊôÓÚ»¹Ô­²úÎïµÄÊÇ__________¡£

(2)¼ÓÈëMnCO3¡¢Zn2(OH)2CO3µÄ×÷ÓÃÊÇ_____________________________________

________________________________________________________________________¡£

¢òÐèÒª¼ÓÈȵÄÔ­ÒòÊÇ____________________________________________________¡£

CµÄ»¯Ñ§Ê½ÊÇ________________________¡£

(3)¸ÃÉú²úÖгýµÃµ½MnO2ºÍZnÒÔÍ⣬»¹¿ÉµÃµ½µÄ¸±²úÆ·ÊÇ________¡£

(4)Èç¹û²»¿¼ÂÇÉú²úÖеÄËðºÄ£¬³ý¿óʯÍ⣬Ð蹺ÂòµÄ»¯¹¤Ô­ÁÏÊÇ________¡£

(5)Òª´ÓNa2SO4ÈÜÒºÖеõ½Ã¢Ïõ(Na2SO4·10H2O)£¬Ðè½øÐеIJÙ×÷ÓÐÕô·¢Å¨Ëõ¡¢________¡¢¹ýÂË¡¢Ï´µÓ¡¢¸ÉÔïµÈ¡£

(6)´ÓÉú²úMnO2ºÍZnµÄ½Ç¶È¼ÆË㣬ÈíÃÌ¿óºÍÉÁп¿óµÄÖÊÁ¿±È´óÔ¼ÊÇ________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


̼ԪËصĵ¥ÖÊÓжàÖÖÐÎʽ£¬ÏÂͼÒÀ´ÎÊÇC60¡¢Ê¯Ä«ºÍ½ð¸ÕʯµÄ½á¹¹Í¼£º

 

 »Ø´ðÏÂÁÐÎÊÌ⣺

    £¨1£©½ð¸Õʯ¡¢Ê¯Ä«¡¢C60.̼ÄÉÃ׹ܵȶ¼ÊÇ̼ԪËصĵ¥ÖÊÐÎʽ£¬ËüÃÇ»¥Îª_____________¡£

    £¨2£©½ð¸Õʯ¡¢Ê¯Ä«Ï©£¨Ö¸µ¥²ãʯī£©ÖÐ̼ԭ×ÓµÄÔÓ»¯ÐÎʽ·Ö±ðΪ____¡¢____¡£

    £¨3£©C60ÊôÓÚ____¾§Ì壬ʯīÊôÓÚ____¾§Ìå¡£

    £¨4£©Ê¯Ä«¾§ÌåÖУ¬²ãÄÚC-C¼üµÄ¼ü³¤Îª142 pm£¬¶ø½ð¸ÕʯÖÐC-C¼üµÄ¼ü³¤Îª154 pm¡£ÆäÔ­ÒòÊǽð¸ÕʯÖÐÖ»´æÔÚC-C¼äµÄ____¹²¼Û¼ü£¬¶øʯī²ãÄÚµÄC-C¼ä²»½ö´æÔÚ____¹²¼Û¼ü£¬»¹ÓÐ____¼ü¡£

    £¨5£©½ð¸Õʯ¾§°ûº¬ÓÐ____¸ö̼ԭ×Ó¡£Èô̼ԭ×Ӱ뾶Ϊr£¬½ð¸Õʯ¾§°ûµÄ±ß³¤Îªa£¬¸ù¾ÝÓ²Çò½Ó´¥Ä£ÐÍ£¬Ôòr= ______a£¬ÁÐʽ±íʾ̼ԭ×ÓÔÚ¾§°ûÖеĿռäÕ¼ÓÐÂÊ____£¨²»ÒªÇó¼ÆËã½á¹û£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


½ðÊôÄø¼°Æ仯ºÏÎïÔںϽð²ÄÁÏÒÔ¼°´ß»¯¼ÁµÈ·½ÃæÓ¦Óù㷺¡£

(1)»ù̬NiÔ­×ӵļ۵ç×Ó(ÍâΧµç×Ó)ÅŲ¼Ê½Îª¡¡¡¡¡¡¡¡

(2)½ðÊôÄøÄÜÓëCOÐγÉÅäºÏÎïNi(CO)4,д³öÓëCO»¥ÎªµÈµç×ÓÌåµÄÒ»ÖÖ·Ö×ÓºÍÒ»ÖÖÀë×ӵĻ¯Ñ§Ê½¡¡¡¡¡¡¡¡¡¢¡¡¡¡¡¡¡¡

(3)ºÜ¶à²»±¥ºÍÓлúÎïÔÚNi´ß»¯Ï¿ÉÓëH2·¢Éú¼Ó³É·´Ó¦¡£

Èç¢ÙCH2=CH2¡¢¢ÚHC¡ÔCH¡¢¢Û¡¢¢ÜHCHO,ÆäÖÐ̼ԭ×Ó²ÉÈ¡sp2ÔÓ»¯µÄ·Ö×ÓÓС¡¡¡¡¡¡¡(ÌîÎïÖÊÐòºÅ),HCHO·Ö×ÓµÄÁ¢Ìå½á¹¹Îª¡¡¡¡¡¡¡¡ÐÎ; 

(4)Ni2+ºÍFe2+µÄ°ë¾¶·Ö±ðΪ69 pmºÍ78 pm,ÔòÈÛµãNiO¡¡¡¡¡¡¡¡FeO(Ìî¡°<¡±»ò¡°>¡±); 

(5)½ðÊôÄøÓëïç(La)ÐγɵĺϽðÊÇÒ»ÖÖÁ¼ºÃµÄ´¢Çâ²ÄÁÏ,Æ侧°û½á¹¹Ê¾ÒâͼÈçÏÂͼËùʾ¡£¸ÃºÏ½ðµÄ»¯Ñ§Ê½Îª¡¡¡¡¡¡¡¡

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


¹¤ÒµÉϱºÉÕÃ÷·¯[ KAl(SO4)2·12H2O]µÄ»¯Ñ§·½³ÌʽΪ£º

4 KAl(SO4)2·12H2O+3S2K2SO4+2Al2O3 +9SO2 +48H2O£¬ÏÂÁÐÅжÏÕýÈ·µÄÊÇ

A£®ÔÚ±ºÉÕÃ÷·¯µÄ·´Ó¦ÖУ¬»¹Ô­¼ÁÓëÑõ»¯¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈÊÇ3£º4

B£®×îºóµÃµ½µÄK2SO4ÈÜÒº³ÊÖÐÐÔ£¬ËùÒÔc(K+)=c(SO42-)

C£®±ºÉÕ²úÉúµÄSO2¿ÉÓÃÓÚÖÆÁòËá,±ºÉÕ948 tÃ÷·¯(M= 474 g/mol)£¬ÈôSO2µÄÀûÓÃÂÊΪ96%£¬¿ÉÉú²úÖÊÁ¿·ÖÊýΪ98%µÄÁòËá432 t

D£®¹¤ÒµÉÏÒ±Á¶Al2O3ÖƵÃAl£¬ÒÔAlºÍNiO(OH)Ϊµç¼«£¬NaOHÈÜҺΪµç½âÒº×é³ÉÒ»ÖÖÐÂÐ͵ç³Ø£¬·ÅµçʱNiO(OH)ת»¯ÎªNi(OH)2£¬¸Ãµç³Ø·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ

Al£«3NiO(OH)£«NaOH   NaAlO2£«3Ni(OH)2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º


12.4 g Na2Rº¬Na£«0.4 mol£¬ÔòNa2RµÄĦ¶ûÖÊÁ¿Îª________£¬RµÄÏà¶ÔÔ­×ÓÖÊÁ¿Îª________¡£º¬RµÄÖÊÁ¿Îª1.6 gµÄNa2R£¬ÆäÎïÖʵÄÁ¿Îª________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸